Problem: The equation of a circle $C$ is $x^2+y^2-8x+16y+31 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2-8x) + (y^2+16y) = -31$ $(x^2-8x+16) + (y^2+16y+64) = -31 + 16 + 64$ $(x-4)^{2} + (y+8)^{2} = 49 = 7^2$ Thus, $(h, k) = (4, -8)$ and $r = 7$.